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Re: Weird trigonometric integral and Simplification question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg31789] Re: [mg31777] Weird trigonometric integral and Simplification question
*From*: BobHanlon at aol.com
*Date*: Sun, 2 Dec 2001 04:25:05 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
In a message dated 2001/12/1 4:50:18 AM, bta at attewode.com writes:
>Simplification question
>
> If I set lambda=(1+Sqrt[5])/2, and then ask Mathematica to
>Expand[lambda^7] or ask for any other power of lambda, Mathematica
>gives a nice result that I can easily rewrite by hand in the form a + b
>lambda where a and b are integers. Is there any way to get Mathematica
>to
>write the result in this form? (I'm not interested in the answer involving
>the Fibonacci sequence, I'm interested in learning how to manipulate
>expressions in Mathematica.)
Once you give lambda a value you cannot express anything literally in
terms of lambda since Mathematica will substitute the assigned value.
However, the symbol GoldenRatio which is equivalent to your lambda
can be used
{GoldenRatio, FunctionExpand[GoldenRatio]}
{GoldenRatio, (1/2)*(1 + Sqrt[5])}
Clear[f];
f[n_Integer?NonNegative] :=
Module[{a,b, lambda = (1+Sqrt[5])/2, soln1, soln2},
Off[Solve::svars];
soln1 = FullSimplify[Solve[a+b*lambda == lambda^n, {a,b}][[1]]];
soln2 =
Solve[Coefficient[FullSimplify[a /. soln1 /. Sqrt[5] -> x], x] == 0,
b][[1]];
On[Solve::svars];
Simplify[a+b*GoldenRatio /. Join[soln1 /. soln2, soln2]]];
f[7]
8 + 13*GoldenRatio
Table[{GoldenRatio^n,"==", t=f[n],
FullSimplify[GoldenRatio^n - t]\[Equal]0}, {n, 0, 7}]//TableForm
>
> Similarly if I ask Mathematica for Inverse[{{a,b},{c,d}}] it gives
>a
>nice result, but how do I get it to factor out a d-b c?
>
inv = Inverse[{{a,b},{c,d}}]
If you factor something out of a matrix, Mathematica will immediately
carry out the multiplication or division. You can use an undefined
operator to delay the operations
CircleTimes[(t=Denominator[inv[[1,1]]]) * inv, 1/t]
inv == Times@@%
Bob Hanlon
Chantilly, VA USA
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